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1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
a. \(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}\)
= \(\left(\dfrac{5}{7}+\dfrac{-12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)\)
= \(\left(-1\right)+1=0\)
b. \(\left(7-2\dfrac{3}{5}\right).2\dfrac{4}{33}-2\dfrac{3}{5}:\dfrac{1}{2}+\dfrac{7}{5}\)
= \(\dfrac{22}{5}.2\dfrac{4}{33}-\dfrac{26}{5}+\dfrac{7}{5}\)
= \(\dfrac{28}{3}-\dfrac{33}{5}=\dfrac{41}{15}\)
c\(\dfrac{-7}{9}:\dfrac{8}{15}+\dfrac{-7}{9}.\dfrac{7}{15}+5\dfrac{7}{9}\)
= \(\dfrac{-56}{135}+\dfrac{-49}{135}+5\dfrac{7}{9}\)
= \(\dfrac{-7}{9}+5\dfrac{7}{9}=5\)
d. \(\left(16\dfrac{3}{8}-19\dfrac{3}{4}\right)-\left(12\dfrac{3}{8}-17\dfrac{3}{4}\right)\)
= \(16\dfrac{3}{8}-19\dfrac{3}{4}-12\dfrac{3}{8}+17\dfrac{3}{4}\)
= \(\left(16\dfrac{3}{8}-12\dfrac{3}{8}\right)-\left(19\dfrac{3}{4}-17\dfrac{3}{4}\right)\)
=\(4+2=6\)
a)\(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}\)
=\(\left(\dfrac{5}{7}-\dfrac{12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)=-1+1=0\)
b)\(\left(7-2\dfrac{3}{5}\right).2\dfrac{4}{33}-2\dfrac{3}{5}:\dfrac{1}{2}+\dfrac{7}{5}\)
=\(\dfrac{22}{5}.\dfrac{70}{33}-\dfrac{13}{5}.2+\dfrac{7}{5}\)
=\(\dfrac{28}{3}-\dfrac{26}{5}+\dfrac{7}{5}=\dfrac{83}{15}\)
Các câu sau tương tự
a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)
b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c) \(=-\dfrac{80}{9}\)
Bài 1:
a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{2}{4}\)
\(=\dfrac{3}{4}\)
b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
\(=\dfrac{1}{2}+\dfrac{4}{5}\)
\(=\dfrac{5}{10}+\dfrac{8}{10}\)
\(=\dfrac{9}{5}\)
c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)
\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)
\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)
\(=\dfrac{7}{3}+\dfrac{28}{3}\)
\(=\dfrac{35}{3}\)
d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)
\(=\dfrac{1}{6}-\dfrac{7}{2}\)
\(=\dfrac{1}{6}-\dfrac{21}{6}\)
\(=\dfrac{-10}{3}\)
e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)
\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\dfrac{2}{3}\)
f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{3}{2}\)
\(=\dfrac{2}{2}=1\)
g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)
\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)
\(=\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{2}{4}-\dfrac{3}{4}\)
\(=\dfrac{-1}{4}\)
h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{9}{28}\)
\(=\dfrac{196}{140}-\dfrac{45}{140}\)
\(=\dfrac{151}{140}\)
i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)
\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)
\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)
\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)
k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
a) \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{4}{-9}+\dfrac{7}{15}\)
=\(\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)+\dfrac{-2}{11}\)
=\(\left(-1\right)+1+\dfrac{-2}{11}\)
=\(\dfrac{-2}{11}\)
b) \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)
=\(\dfrac{-5}{12}+\dfrac{6}{11}+\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\)
=\(\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\dfrac{7}{17}\)
=\(0+0+\dfrac{7}{17}\)
=\(\dfrac{7}{17}\)
c) A= \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
A=\(49\dfrac{8}{23}-5\dfrac{7}{32}-14\dfrac{8}{23}\)
A=\(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)-5\dfrac{7}{32}\)
A=\(35-5\dfrac{7}{32}\)
A=\(35-\dfrac{167}{32}=\dfrac{953}{32}\)
d) C=\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)
C=\(\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}.1+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}+\dfrac{17}{7}=2\)
a, `(-5)/9+8/15+(-2)/11+4/(-9)+7/15`
`=-5/9+8/15-2/11-4/9+7/15`
`=(-5/9-4/9)+(8/15+7/15)-2/11`
`=-9/9+15/15-2/11`
`=-1+1-2/11`
`=-2/11`
b, `((-5)/12+6/11)+(7/17+5/11+5/12)`
`=-5/12+6/11+7/17+5/11+5/12`
`=(-5/12+5/12)+(6/11+5/11)+7/17`
`=0+11/11+7/17`
`=1+7/17`
`=17/17+7/17`
`=24/17`
c, `A=49 8/23 - (5 7/32 + 14 8/23)`
`A=49 8/23 - 5 7/32 - 14 8/23`
`A=(49 8/23 - 14 8/23)-5 7/32`
`A=35 - 167/32`
`A=953/32`
d, `C=(-3)/7.5/9+4/9.(-3)/7+2 3/7`
`C=-3/7 . 5/9-4/9 . 3/7+17/7`
`C=-3/7.(5/9+4/9)+17/7`
`C=-3/7 . 1+17/7`
`C=2`
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
a: \(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)
\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{122}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)
b: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}\)
\(=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)
\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)
\(=\dfrac{-35\cdot5-11}{31}=\dfrac{-186}{31}=-6\)
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)